# The Stacks Project

## Tag: 00HS

This tag has label algebra-lemma-flat-going-down and it points to

The corresponding content:

Lemma 9.36.17. Let $R \to S$ be flat. Let $\mathfrak p \subset \mathfrak p'$ be primes of $R$. Let $\mathfrak q' \subset S$ be a prime of $S$ mapping to $\mathfrak p'$. Then there exists a prime $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p$.

Proof. Namely, consider the flat local ring map $R_{\mathfrak p'} \to S_{\mathfrak q'}$. By Lemma 9.36.16 above this is faithfully flat. By Lemma 9.36.15 there is a prime mapping to $\mathfrak p R_{\mathfrak p'}$. The inverse image of this prime in $S$ does the job. $\square$

\begin{lemma}
\label{lemma-flat-going-down}
Let $R \to S$ be flat. Let $\mathfrak p \subset \mathfrak p'$
be primes of $R$. Let $\mathfrak q' \subset S$ be a prime of $S$
mapping to $\mathfrak p'$. Then there exists a prime
$\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p$.
\end{lemma}

\begin{proof}
Namely, consider the flat local ring map
$R_{\mathfrak p'} \to S_{\mathfrak q'}$.
By Lemma \ref{lemma-local-flat-ff} above this is faithfully
flat. By Lemma \ref{lemma-ff-rings} there is a prime mapping to
$\mathfrak p R_{\mathfrak p'}$. The inverse image of this
prime in $S$ does the job.
\end{proof}


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